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This is related to my previous question. I assume that the two wires have the same charge density with the same sign ($\lambda_1 = \lambda_2 = \lambda$)

If I want to take the same approach, the potential is still the same,

$V = k\lambda \log(1/r)$.

The energy is, I think, the integral over the (second) wire, since this is the direction where I add charges. Since the wires are paralleled, $r$ is constant in the integral

$U = \int_{-\infty}^{\infty} l\lambda^2 \log(1/r) dx$

Which explodes.

I actually expect this, since I keep adding charges at a finite distance from the other wire.

On the other hand though, I think there are well defined solutions for the energy of infinite area capacitors which are worse. Or is it only the fields?

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1 Answer 1

You are trying to calculate the total energy in the electric field, so it's no surprise that it's infinite, since your system is infinite. More meaningful would be the energy density $\mathcal{E}$, or the electric field $E \sim \mathcal{E}^{1/2}$. Same goes for the infinite capacitor, the total energy is infinite, but the electric field at any given point between the plates (and hence the energy density at that point) is finite.

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